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UPSC NDA 2 2019 Mathematics Solved Paper

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Question Numbers: 84-85

Directions: Read the following information and answer the two items that follow:

Consider the equation

x^{y} = e^{x-y}

}

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Question : 85 of 120

Marks: +1, -0

What is

\frac{d^{2}y}{dx^{2}}

at x = 1 equal to?

0
1
2
4

Solution:

Concept:

Differentiate

\frac{dy}{dx}

using quotient rule to get

\frac{d^{2}y}{dx^{2}}

, then substitute

x=1

.

Explanation:

We are given

\frac{dy}{dx} = \frac{\ln x}{(1+\ln x)^{2}}

.

Use quotient rule:

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}

.

Let

u = \ln x

,

v = (1+\ln x)^{2}

.

\frac{du}{dx} = \frac{1}{x}

.

\frac{dv}{dx} = 2(1+\ln x) \cdot \frac{1}{x}

by chain rule.

Thus

\frac{d^{2}y}{dx^{2}} = \frac{(1+\ln x)^{2} \cdot \frac{1}{x} - \ln x \cdot 2(1+\ln x) \cdot \frac{1}{x}}{(1+\ln x)^{4}}

.

Factor

\frac{(1+\ln x)}{x}

in numerator:

\frac{\frac{(1+\ln x)}{x}[(1+\ln x) - 2\ln x]}{(1+\ln x)^{4}} = \frac{1 - \ln x}{x(1+\ln x)^{3}}

.

At

x=1

,

\ln 1 = 0

. Numerator

=1-0=1

, denominator

=1 \cdot (1+0)^{3}=1

.

Hence

\frac{d^{2}y}{dx^{2}} = 1

at

x=1

.

Answer:

1

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